Description
We give the following inductive definition of a “regular brackets” sequence:
- the empty sequence is a regular brackets sequence,
- if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
- if a and b are regular brackets sequences, then ab is a regular brackets sequence.
- no other sequence is a regular brackets sequence
For instance, all of the following character sequences are regular brackets sequences:
(), [], (()), ()[], ()[()]
while the following character sequences are not:
(, ], )(, ([)], ([(]
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1< i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence.
Given the initial sequence ([([]])]
, the longest regular brackets subsequence is [([])]
.
Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (
, )
, [
, and ]
; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.
Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
Sample Input
((()))
()()()
([]])
)[)(
([][][)
end
Sample Output
6
6
4
0
6
题意:现在已知一个由'()''[]'组成的括号序列 问其中括号匹配数最大的子序列个数
思路:
这里的状态转移是以一个if为基础的,如果s[i]与s[j]匹配,那么明显的dp[i][j] = dp[i+1][j-1];然后在这个基础上枚举分割点k.
状态转移方程:dp[i][j]表示第i~j个字符间的最大匹配对数。
if(s[i] 与 s[j]匹配) dp[i][j] = d[[i+1][j-1] +1;
dp[i][j] = max(dp[i][j],dp[i][k]+dp[k+1][j]);
最后乘2即可
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> #define ll long long int using namespace std; inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1}; int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1}; const int inf=0x3f3f3f3f; const ll mod=1e9+7; int dp[107][107]; string s; int jug(int i,int j){ return (s[i]=='['&&s[j]==']')||(s[i]=='('&&s[j]==')'); } int main(){ ios::sync_with_stdio(false); while(cin>>s){ if(s=="end") break; memset(dp,0,sizeof(dp)); int n=s.length(); for(int len=2;len<=n;len++){ for(int i=1;i+len<=n+1;i++){ int j=i+len-1; dp[i][j]=dp[i+1][j-1]+jug(i-1,j-1); for(int k=i;k<j;k++) dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]); } } cout<<2*dp[1][n]<<endl; } return 0; }